Engineering
Physics
Conduction of Heat
Question
Three rods made of the same material and having same cross-sectional area but different lengths 10cm, 20 cm and 30 cm are joined as shown. The temperature of the joint is:
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Solution
Using Junction Rule:
\(\frac{T-20}{20.K/A}+\frac{T-10}{10.K/A}+\frac{T-30}{30.K/A}=0\)
\(T=16.4^\circ C\)